To analyze and represent such signals we need wavelets that are local in space and frequency. Typically this is achieved by building wavelets which have compact support (localization in space), which are smooth (decay towards high frequencies), and which have _vanishing moments_ (decay towards low frequencies).

I understand that compact support means the function is non-zero on a limited portion of its domain. "Smoothness" I believe means that there are no corners in the basis function, ie it would not take infinitely high frequencies to represent the curve (using Fourier analysis).

I do not know what a "vanishing moment" means, intuitively, though. Why would you call a "decay to towards low frequencies" a vanishing moment?

2 Answers
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The basic idea is that a wavelet has $p$ vanishing moments if and only if the wavelet scaling function can generate polynomials up to degree $p-1$. The "vanishing" part means that the wavelet coefficients are zero for polynomials of degree at most $p-1$, that is, the scaling function alone can be used to represent such functions. More vanishing moments means that the scaling function can represent more complex functions. Loosely, you can think of it as

$$\textrm{more vanishing moments } \rightarrow \textrm{ complex functions can be represented with a sparser set of wavelet coefficients.}$$

The "moments" part comes from the fact that this is all equivalent to saying that the first $p$ derivatives of the Fourier transform of the wavelet filter all are zero when evaluated at 0. This is perfectly analogous to the probabilistic idea of a "moment generating function" of a random variable, which is basically the Fourier transform, and the $n$-th derivative evaluated at zero gives the $n$-th moment of the variable (i.e. the expected value, the expected value of the square, of the cube, etc.) So these Fourier transform derivative-zeros correspond to integrals back in the time/space domain that must be zero for the wavelet. In a sense, these conditions mean that the wavelet is "unbiased." It doesn't skew the function that is being transformed because the wavelet itself has no expected effect on a function until that function has a non-trivial $p+1$ order derivative.

Added: Section 5.2.1 at this link shows the integrals that I'm referring to and, I think, does a good job illustrating why you might refer to this as a "decay toward infinity" kind of property.

A couple questions: 1) Is the link broken? 2) When you say "equivalent to saying that the first p derivatives of the Fourier transform of the wavelet filter all are zero when evaluated at 0.", do you mean evaluated at zero frequency? Isnt that then just the dot-product of the wavelet with the function in question?...
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MohammadNov 13 '12 at 16:27

"When wavelet-functions coefficients are expressed as z-transform, then the number of zeros at pi correspond to the number of vanishing moments. So p zeros in pi means p vanishing moments." dsprelated.com/showmessage/46547/1.php
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endolithMar 6 '13 at 18:40

The n-th moment of a function is equal to the n-th derivative of of its fourier transform at zero frequency. So more the number of zero moments, more the number of zero derivatives and smoother the signal decays from mid freqeuncy to dc in the frequency domain